\chapter{Improper Integrals}

An \index{improper integral} improper integral is the area of a shape that extends out to infinity in
some direction.  In many cases, such a shape has infinite area, and we say
that the improper integral is $\infty$.  However, in certain cases the
total area is finite, because the part extending out infinitely far is
``pinched closely together,'' so that the very narrow part going out
infinitely far does not contribute much area.  This all sounds rather
vague.  The best way to understand improper integrals is through examples.

\begin{example}
\label{example:9.1}

Find the area under the reciprocal cube curve $y=1/x^3$ to
the right of the line $x=1$.

\smallskip
First, suppose that we bound the area on the right at $x=b$, where $b$ is a
very large upper limit of integration.  Then we compute 
$$
\int_1^b\frac{dx}{
x^3}=-\frac{1}{ 2 }x^{-2}\bigm|_1^b=\frac{\left(1-b^{-2}\right) }{2}.
$$  
For example, the area from $x=1$ to $x=100$ is
exactly $0.49995$, the area from $x=1$ to $x=1000000$ is exactly
$0.4999999999995$, and so on.  Clearly, the limit as
$b\longrightarrow\infty$ is $0.5$.  This is what we mean by the \index{area out to
infinity} area out to
infinity: 
$$
\int_1^\infty\frac{dx}{x^3}
=\lim_{b\longrightarrow\infty}\int_1^b\frac{dx}{x^3}
=\lim_{b\longrightarrow\infty}\frac{\left(1-b^{-2}\right)}{2}=0.5.
$$
\end{example}

\begin{example} 
\label{example:9.2}

Find the area under the reciprocal curve $y=1/x$ to the
right of the line $x=1$.

\smallskip

\begin{rightindent}{2.5 truein}

 
Here we have 
$$
\lim_{b\longrightarrow\infty}\int_1^b\frac{dx}{ x}=
\lim_{b\longrightarrow\infty}\ln b=\infty.
$$ 
Roughly speaking, the reciprocal cube function pinches together with the
$x$-axis a lot faster than does the reciprocal function, and so we get
finite area with $1/x^3$ but not with $1/x$.
\end{rightindent}
\hfill
%-figure JA -----------------------------
\begin{psfigure}{3in}{1.75in}{figJA.ps}
%\putcrosshair{0,0}
\puttext{5, 12}{\Small$y = 1/x$}
\puttext{-35, -5.}{\Small$y = 1/x^3$}
\end{psfigure}
%------------------------------

\end{example}

\begin{example}
\label{example:9.3}

Let $y=1/x^n$, where $n$ is a positive constant not equal
to 1.  Find the area under this curve to the right of the line $x=1$.

\smallskip
Here we want the limit as $b\longrightarrow\infty$ of
$\int_1^bx^{-n}dx=\frac{1}{ -n+1}x^{-n+1}\bigm|_1^b=\frac{1}{ n-1}\left(1-
b^{1-n}\right)$.  What happens as $b\longrightarrow\infty$ depends on the
exponent of $b$, which is $1-n$.  If this is positive, i.e., if $n<1$, then
the limit is infinite.  On the other hand, if $n>1$, then $b^{1-n}$
approaches 0, and so we obtain the finite area $1/(n-1)$.  We can summarize
the situation as follows (note that in Example~\ref{example:9.2} we saw that the area is
infinite in the special case $n=1$): 
$$
\int_1^\infty\frac{dx}{ x^n}
=
\begin{cases}\frac{1}{ n-1}& \text{if $n>1$;}\\
~~&\\
\infty& \text{if $n \le 1$.}
\end{cases}
$$
\end{example}


\begin{example} 
\label{example:9.4}

Find the area bounded by the $x$- and $y$-axes, the line
$x=1$, and the reciprocal cube root curve $y=1/{\root 3 \of x}$.

\smallskip

\begin{rightindent}{3.0 truein}
Here the region extends out to infinity in the $y$-direction.  But the
curve comes very close to the $y$-axis (its asymptote), so there is very
little area enclosed when $y$ is large.  For this reason, as we shall see,
the total area is finite.  The area is the improper integral
$\int_0^1x^{-1/3}dx$.  To evaluate this, let us compute the ordinary ({\it
not} improper) integral $\int_a^1x^{-1/3}dx$ for $a>0$, and then see what
happens when $a\longrightarrow 0$.  That is,
\end{rightindent}
\hfill
%-figure JB -----------------------------
\begin{psfigure}{1.8in}{1.8in}{figJB.ps}
%\putcrosshair{0,0}
\puttext{-16.7, 13.9}{\Small $y = 1/\sqrt[3]{x}$}
\puttext{13.9, -27.8}{\Small $y = 1/x^3$}
\end{psfigure}
%------------------------------
$$
\int_0^1x^{-1/3}dx=\lim_{a\longrightarrow 0}\int_a^1x^{-1/3}dx=
\lim_{a\longrightarrow 0}\left(\frac{3}{ 2}x^{2/3}\bigm|^1_a\right)=
\lim_{a\longrightarrow 0}\left(\frac{3}{ 2}\left(1-a^{2/3}\right)\right)=3/2,
$$
because $a$ raised to any positive power approaches 0 as $a\longrightarrow
0$.

\end{example}

\medskip

In Example~\ref{example:9.4}, suppose we remove from our region the square
of unit side that is bounded by the $x$- and $y$-axes, the line $x=1$, and
the line $y=1$.  The remaining area is the area to the left of the curve
$x=1/y^3$ above the line $y=1$.  This is exactly the same area as in
Example~\ref{example:9.1}, except that the roles of $x$ and $y$ are
reversed.  Thus, we expect the area in Example~\ref{example:9.4} to equal 1
plus the area in Example~\ref{example:9.1}.  Thus, the value of 3/2 in the
last paragraph agrees with the value of 1/2 in Example~\ref{example:9.1}.

\smallskip

In the same way as in Example~\ref{example:9.4}, we can compute the area
bounded by the $x$- and $y$-axes, the line $x=1$, and the curve $y=1/x^n$.
The computation is just like the special case when $n=1/3$.  The result is:
$$
\int_0^1\frac{dx}{ x^n}=
\begin{cases}\frac{1}{ 1-n}& \text{if $n<1$;}\\
~~&\\
\infty& \text{if $n\ge 1$.}
\end{cases}
$$


\begin{example}
\label{example:9.5}
Find $\int_{-2}^2dx/\sqrt{4-x^2}$.

\smallskip

Here the integrand $(4-x^2)^{-1/2}$ is infinite at {\it both} endpoints of
the interval of integration --- the region goes out to infinity along the
line $x=-2$ and along the line $x=2$.  To get a region that does not extend
to infinity, we integrate between $-a$ and $a$, where $a$ is a little less
than $2$.  Then we take the limit as $a\longrightarrow 2$.  In evaluating
the integral, we use Example~\ref{example:7.8} (with $r$ set equal to 2) of
the section on the Arcsin.  Thus,
\begin{align*}
\int_{-2}^2\frac{dx}{\sqrt{4-x^2}}&=
\lim_{a\longrightarrow 2}\int_{-a}^a\frac{dx}{\sqrt{4-x^2}}
=\lim_{a \longrightarrow 2}\arcsin(x/2)\bigm|_{-a}^a\\
&=\arcsin(1)-  \arcsin(-1)=\frac{\pi}{ 2}-\left(-\frac{\pi}{ 2}\right)=\pi.
\end{align*}
\end{example}


\begin{example}
\label{example:9.6}\index{a solid with finite volume but infinite surface area}
The region in the fourth quadrant bounded by the $y$-axis,
the line $y=-1$, and the curve $y=-1/x$ is rotated around the $y$-axis to
obtain a container (an infinitely deep funnel).  (a) Find the volume of the
funnel.  (b) Show that the funnel has infinite surface area.  Thus, one
could fill the funnel entirely with water, but if one cuts the side of
the funnel into small pieces and spreads them out on a table, there is
no table of finite size that would hold all of the pieces!

\smallskip

(a) We first write $x$ as a function of $y$: $x=-1/y$.  Then in the volume
integral $\pi\int_{-\infty}^{-1}dy/y^2$, we make the change of variables
$u=-y$, so that $du=-dy$.  This transforms the volume to
$-\pi\int_\infty^1dy/y^2=\pi\int_1^\infty du/u^2$ (we used the fact that
the sign of a definite integral changes when you reverse the upper and
lower limits of integration).  The last integral was computed in Example 9.3
(take $n=2$): it has finite value 1.  Thus, the volume of the funnel is
$\pi$.

\smallskip

(b) We have not developed the formula for the surface area of a volume of
revolution.  However, we can answer the question in part (b) by observing
that the surface area is greater than the area of the flattened surface
contained in the third and fourth quadrants below $y=-1$ between the curves
$x=1/y$ and $x=-1/y$.  This flat area is equal to
$\int_{-\infty}^{-1}\frac{-2}{ y}dy=2\int_1^\infty du/u$ (where we make the
same substitution $u=-y$ as before).  But the latter improper integral is
infinite, as we saw in Example~\ref{example:9.2}.  Therefore, the funnel has infinite
surface area.

\end{example}

\begin{example}
\label{example:9.7} \index{freeway problem}
Determine how the noise from a freeway drops off as you
move farther away.  Make the following simplifying assumptions: (1) the
amount of noise generated by a segment of the freeway is proportional to
the segment's length (i.e., the traffic is uniformly distributed along the
freeway); (2) the contribution of a small segment of the freeway to the
amount of noise you hear is inversely proportional to the square of your
distance from that segment; (3) the freeway is a straight line extending
infinitely far in both directions.  (We are also assuming that the noise
from different parts of the freeway add up without ``canceling each
other,'' and we are using a linear scale of noise, rather than a
logarithmic scale such as the decibel system.)


\begin{centering}
%-figure JC -----------------------------
\begin{psfigure}{4in}{1.5in}{figJC.ps}
%\putcrosshair{0,0}
\puttext{0, 20.}{\Small$0$}
\puttext{-125., 5.}{\Small$\longleftarrow\,-\infty$}
\puttext{125., 5.}{\Small$\infty\,\longrightarrow$}
\puttext{-5., 0}{\Small$r$}
\puttext{35, -5}{\Small$\sqrt{r^2+x^2}$}
\puttext{36, 30}{\Small$x$}
\put(36,25){\vector(0,-1){7.5}}
\puttext{60., 30}{\Small$x+\Delta x$}
\put(60,25){\vector(-2,-1){17.5}}
\end{psfigure}\\
\end{centering}

Let $r$ be our perpendicular distance from the freeway.  Take the freeway
to be the $x$-axis, with 0 located at the point closest to us.  We shall
find the total noise by the same type of procedure that we used to find
definite integral formulas for the area under a curve, the volume of a
solid of revolution, and the total moment of a uniform plate.  In those
cases we had the integral from $a$ to $b$ of the contribution (to area,
volume, moment, or whatever) coming from the part between $x$ and $x+\Delta
x$ (where we replace $\Delta x$ by $dx$ when we go from the \index{Riemann sum} Riemann sum to
the integral).  

For the sake of review, the chart on the next page shows
the contribution from the part between $x$ and $x+\Delta x$
and the integral that resulted.

\medskip
\begin{small}

\begin{centering}
\begin{tabular}{||l|l|l||} \hline\hline
TYPE OF PROBLEM &CONTRIBUTION& FORMULA\\ \hline
&&\\
area~under~curve  & $\approx f(x)\Delta x$       & $\displaystyle \int_a^b f(x)dx$\\
%&&\\
%\hline
%&&\\
vol~of~revolution & $\approx\pi f(x)^2\Delta x$  &
$\displaystyle\pi\int_a^b f(x)^2dx$\\
%&&\\
%\hline
%&&\\
moment~of~plate   & $\approx D\,x\,f(x)\Delta x$ & $\displaystyle
D\int_a^bx\,f(x)dx$\\
&&\\
\hline\hline
\end{tabular}\\
\end{centering}

\end{small}

\medskip


So in the freeway problem we want to find the contribution to noise that comes
from the tiny segment $[x,x+\Delta x]$.  According to assumptions (1) and
(2), the noise
you hear from this segment is proportional to the length $\Delta x$ of the
segment and inversely proportional to the square of your distance from the
segment.  In computing this distance, we take the distance from the left
endpoint of the segment (since $\Delta x$ is very small, this is very close
to the distance from any other point on the segment); by the \index{Pythagorean
theorem} Pythagorean
theorem, the square of this distance is $r^2+x^2$.  Thus, if we let $K$
denote the constant of proportionality, we find that the contribution is
$$
\approx K\frac{\Delta x}{ r^2+x^2}.
$$ 


\smallskip

\noindent

\medskip

Thus, the noise coming from the entire highway between $a$ and $b$ is
found by summing up these contributions --- or rather (after taking
the limit as $\Delta x\longrightarrow 0$) taking the integral with
$\Delta x$ replaced by $dx$:  Noise$=\int_a^b\frac{K\,dx}{ r^2+x^2}$.  Thus,
\begin{align*} 
{\rm Noise}&=K\int_a^b\frac{dx}{ r^2+x^2}
=\frac{K}{ r^2}\int_a^b\frac{dx}{1+(x/r)^2}=\\ 
&=\frac{K}{ r^2}\int_{a/r}^{b/r}\frac{r\,du}{ 1+u^2}
=\frac{K}{ r}\int_{a/r}^{b/r}\frac{du}{ 1+u^2} \\
&=\frac{K}{ r}\arctan\,u \bigm|_{a/r}^{b/r}
=\frac{K}{ r}\biggl(\arctan\bigl(\frac{b}{ r}\bigr)- \arctan\bigl(\frac{a}{ r}\bigr)\biggr).
\end{align*} 
If the freeway had finite length and were contained between $a$ and $b$,
then this would give us a rather complicated expression for the noise from
the freeway.  It would not be easy to say how the noise drops off as our
distance $r$ increases.  However, at this point we use simplifying
assumption (3), which tells us to take the limit as $a\longrightarrow
-\infty$ and $b\longrightarrow\infty$, i.e.,
\begin{align*}
\text{Noise}&=\frac{K}{ r}\biggl(\lim_{b\longrightarrow\infty}
\arctan\bigl(\frac{b}{ r}\bigr)- \lim_{a\longrightarrow-
\infty}\arctan\bigl(\frac{a}{ r}\bigr)\biggr)\\
&=\frac{K}{ r}\biggl(\frac{\pi}{ 2}-\bigl(-\frac{\pi}{ 2}\bigr)\biggr)
=\frac{K\pi}{ r}.
\end{align*}
Since $K$ and $\pi$ are constants, we end up with the simple
conclusion: {\it noise from the freeway drops off proportionally to
distance}.  Thus, if you move twice as far away, the amount of noise drops
only by 50\%.  (We say ``only,'' because in the case of a point source the
noise would drop by a factor of 4, i.e., by 75\%.)

\smallskip

Even though the simplifying assumptions might seem to detract from the
realism of this example, our conclusion roughly agrees with what one
observes in practice.  For example, real estate agents know that noise from
Interstate 5 has a bad effect on property values for quite a distance ---
much more so than would a ``point source'' such as a noisy factory.

\smallskip

Notice the role of simplifying assumption (3).  Although it might seem odd
at first, improper integrals have many practical applications: it often
simplifies the calculations to suppose that something extends out to
infinity.

\end{example}

%
